In this aptitude odd-man-out question, four numbers are given. Three of them are perfect cubes of integers, while one number is not a perfect cube. Among 125, 512, 1321 and 1728, which number is the odd one out because it is not a perfect cube?

Introduction / Context:
This question is a classic odd-man-out problem from the aptitude section, where you are asked to identify which number does not belong to a group based on a hidden numerical property. Here, the relevant concept is the idea of perfect cubes, that is, numbers which can be written in the form n^3 for some integer n. Such questions test your familiarity with basic powers of integers and your ability to quickly recognize standard cube values commonly used in competitive exams.

Given Data / Assumptions:
The four numbers given are 125, 512, 1321 and 1728.
Exactly three of these numbers are perfect cubes of positive integers.
One number is not a perfect cube and must be identified as the odd one out.
No additional operations like addition or subtraction are required; the key is to check the cube property of each number directly.

Concept / Approach:
A perfect cube is any integer that can be written as n^3, where n is an integer. Common cubes that are useful for exams include 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 7^3 = 343, 8^3 = 512, 10^3 = 1000 and 12^3 = 1728. The strategy is to compare the given numbers with known cube values. If a number exactly matches n^3 for some integer n, then it is a perfect cube. If it does not match any such value, it is not a perfect cube and becomes the odd one out.

Step-by-Step Solution:
Step 1: Check 125. We know 5^3 = 5 * 5 * 5 = 125, so 125 is a perfect cube. Step 2: Check 512. We know 8^3 = 8 * 8 * 8 = 512, so 512 is also a perfect cube. Step 3: Check 1728. From standard values, 12^3 = 12 * 12 * 12 = 1728, so this is a perfect cube as well. Step 4: Check 1321. Compare with nearby cubes: 10^3 = 1000, 11^3 = 1331 and 12^3 = 1728. The number 1321 does not match any integer cube value. Step 5: Conclude that 1321 is not a perfect cube, while the other three numbers are perfect cubes of 5, 8 and 12 respectively.
Verification / Alternative check:
An alternative check is to take cube roots (mentally or approximately). The cube root of 125 is exactly 5, of 512 is exactly 8 and of 1728 is exactly 12. However, the cube root of 1321 lies between 10 and 11 and does not give an integer. Since only perfect cubes have integer cube roots, this confirms that 1321 is not a perfect cube. Therefore, it must be the odd number among the four given options.

Why Other Options Are Wrong:
125 is 5^3, so it correctly satisfies the perfect cube property and cannot be the odd one out.
512 is 8^3, another standard cube, so it fits well with the common pattern of perfect cubes.
1728 is 12^3, again a perfect cube frequently used in exam questions, so it clearly belongs to the main group.
Only 1321 fails to match any integer cube and therefore does not belong with the others.

Common Pitfalls:
A common mistake is to confuse squares and cubes, especially when dealing with larger numbers. Some students may only check whether the numbers look familiar instead of verifying their cube roots. Another error is to assume that any number close to a known cube, like 1321 near 1331, might still be considered a cube. In aptitude exams, however, the relationship is exact, so approximate values cannot be treated as perfect cubes. Always verify by relating to known cube values or by checking the approximate cube root.

Final Answer:
The number that is not a perfect cube and is therefore the odd one out is 1321.